Network Theorems MCQ Quiz in தமிழ் - Objective Question with Answer for Network Theorems - இலவச PDF ஐப் பதிவிறக்கவும்

To get the maximum power across the load resistor, the current flows through the load must be maximum.

For R = ∞, it acts as an open circuit and the total source current flows through the load.

Therefore, the maximum power flows through the load at a very high value or infinite value of R.

Maximum power theorem states that for maximum power to be transferred to the load resistance RL, RL must equal the Thevenin Equivalent resistance, i.e.

RL = Rth

But here, we are asked to find the value of R, and not RL that will result in the maximum power to be transferred to load RL. So we cannot go by the standard procedure of equating RL with the Thevenin equivalent resistance.

Concept:

Thevenin's Theorem states that “In any linear, bidirectional circuit having more than one independent source, having more number of active and passive elements, it can be replaced by a single equivalent circuit consisting of equivalent voltage source V_Th and series with equivalent circuit resistance R_Th“.

Where

V_Th = Thevenin Voltage or open circuit voltage

RTh = Thevenin Resistance

For R_th calculation:

All independent sources replaced by their internal resistance (ideal voltage source replaced by a short circuit and ideal current source replaced by an open circuit).

V_th is open circuit voltage 

Calculation:

\({R_{Th}} = \frac{{{r_1} \times {r_2}}}{{{r_1} + {r_2}}} = \frac{{6 \times 6}}{{6 + 6}}\)

R_th = 3 ohm

Applying KVL

-12 + 6I + 6I = 0

12 I = 12

I = 1 A

V_th = 6I

V_th­ = 6 V

Thevenin's Equivalent Circuit

Norton’s theorem is applied to the DC circuits, it will result in a current source with resistance in parallel.

Thevenin’s theorem is applied to the DC circuits, it will result in a voltage source with a resistance in series.

Generally, superposition theorem is used to find voltages and currents. And it is applicable only for linear, bilateral elements.

Superposition theorem is not applicable for power calculation.

Note:

• The superposition theorem states that "in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently."
• The superposition theorem applies only when all the components of the circuit are linear, which is the case for resistors, capacitors, and inductors it is not applicable to networks containing nonlinear elements.

Concept:

Thevenin's Theorem states that “In any linear, bidirectional circuit having more than one independent source, having more number of active and passive elements, it can be replaced by a single equivalent circuit consisting of equivalent voltage source VTh and series with equivalent circuit resistance RTh“.

Where

VTh = Thevenin Voltage or open circuit voltage

RTh= Thevenin Resistance 

For RTh All independent sources replaced by their internal resistance (ideal voltage source replaced by a short circuit and ideal current source replaced by an open circuit.

V_th is open circuit voltage

Calculation:

Now voltage across terminal a - b is

\({V_{ab}} = 10\left( {\frac{{50}}{{50 + 50}}} \right) = 5V\)

The Thevenin’s resistance can be find by neglecting the voltages source.

R_ab = 25 Ω

New Thevenin equivalent network is,

Millman’s Theorem:

It states that – when a number of voltage sources (V1, V2, V3……… Vn) are in parallel having internal resistance (R1, R2, R3………….Rn) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R. 

The equivalent circuit parameter

\(V = \frac{{\frac{{{V_1}}}{{{R_1}}} + \frac{{{V_2}}}{{{R_2}}} + \frac{{{V_3}}}{{{R_3}}}}}{{\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}}}\)

\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

V = equivalent voltage

R = equivalent resistance

This theorem is nothing but a combination of Thevenin’s Theorem and Norton’s Theorem. It is very useful theorem to find out voltage across the load and current through the load. This theorem is also called as parallel generator theorem

The correct answer is  R_n = 1/1.6 Ω , I_sc = 1/2 A.

Solution:

To find Norton resistance shortcircuit the independent voltage source and apply a 1 V test signal and calculate the current through it.

i + 2 I + I = 1 / 10

i + 3 I = 1/10

Also I = - 1 / 2

So i - 3/2 = 1/10

i = 16 / 10 A

∴ i = 1.6 A

R_n =  V / I = 1 / 1.6

∴ R_n = 1/1.6 Ω 

To find Isc, Short circuit the given terminals.

As we can see in the above figure the short circuit voltage across 2Ω resistance is zero hence current in 2 Ω resistance is zero.

Due to that dependent current source is open-circuited because of the current I = 0 A.

I_sc = 5/10 

= 1/2 A

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown:

Vth = Open circuit Voltage at a – b (by removing the load), i.e.

R_th = equivalent resistance obtaining by deactivating sources.(Independent voltage source is short-circuited

and Independent current source is open  circuited)

Calculation:

Calculation of V_th

Apply KVL in the loop

20 - 3I - 6I + 10 = 0

I = (20 + 10) / (3 + 6)

I = 30 / 9

V_6Ω  = (30 / 9) × 6 

V6Ω = 20 V

Now apply KVL in the outer loop

- 10 + V_6Ω  - V_ab = 0

V_ab = V_th = (20 - 10) V = 10 V

where, I = current through loop 

Calculation of R_th

R_th = [3 || 6] + 3

Rth = 2 + 3 = 5 Ω

Concept :

Superposition Theorem

Superposition theorem states that in any linear, active, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance. The superposition theorem is used to solve the network where two or more sources are present and connected.

The internal impedances of Independent sources are replaced as follows

•             Replacing all other independent voltage sources with a short circuit

•             Replacing all other independent current sources with an open circuit

Calculations :

1.When 20 V Source Is Acting Alone

Applying KCL

\(\frac{{{V_x} - 20\;}}{{20}} + \frac{{{V_x}}}{4} = 0.1{V_x}\)

V_x (0.05 + 0.25 - 0.1) = 1

V_x = 5 V

2.When 4 A Source Is Acting Alone

Applying KCL,

0.1V_x + 4 = V_x /4 + V_x / 20

V_x  = 20 V

Total response when both sources acting is the sum of responses when individual sources acting alone

V_x = V_x1 + V_x2 = 5 + 20 = 25 V

Concept:

Norton's Theorem:

In any linear, bidirectional circuit having more than one independent source, having more the active and passive element it can be replaced by a single equivalent current source IN in parallel with an equivalent resistance RN. 

Where 

IN = Norton or short circuit current

RN = Norton's resistance

Procedure in order to find Norton’s equivalent circuit, when only the sources of independent type are present.

•  Consider the circuit diagram by opening the terminals with respect to which, Norton’s equivalent circuit is to be found.
•  Find Norton’s current IN by shorting the two opened terminals of the circuit.
•  Find Norton’s resistance RN across the open terminals of the circuit, eliminating the independent sources present in it. 
•  Norton’s resistance RN will be the same as that of Thevenin’s resistance RTh.
•  Draw Norton’s equivalent circuit by connecting a Norton’s current IN in parallel with Norton’s resistance RN.

Explanation:

Given circuit is 

To find Norton current through terminal ab, ab terminal is short circuited, so no current will flow through the 5 Ω resistor.

Now the circuit will look like

\(6=\frac{V_x -16}{4}+\frac{V_x}{8+8}\)

16 × 6 = 4(V_x - 16) + V_x

5 V_x = 10 × 16

V_x = 32 V

I_N = V_x / 16 = 32 / 16 = 2 A

To find Norton's resistance, source should be replaced with internal resistance, so

• Current source is open circuited.
• Voltage source is short circuited.

The circuit will become

R_N = 5 || (8 + 4+ 8) = 5 || 20 = (5 × 20) / (5 + 20) = 4 Ω 

So the Norton equivalent circuit is